A small particle is given an initial velocity `v_(0)=10 m//s` along the tangent to the brim of a fixed smooth hemisphere bowl of radius `r_(0)=15sqrt(2)m` as shown in the figure. The particle slides on the inner surface and reaches point `B`, a vertical distance `h=15m` below `A` and a distance `r` from the vertical centerline, where its velocity `v` makes an angle `theta` with the horizontal tangent to the bowl through `B`. If `theta=15 K^(@)`. Find the value of `K`. (take `g=10 m//s^(2)`)

Answer

2

Explanation

`t_(DE)=sqrt((2d)/(g))`
`4d=(v_(0))sqrt((2d)/(g))`
`v_(0)^(2)=8gd`
energy - work theorem
`(1)/(2)kdelta^(2)=(mumgd)+mgd+(1)/(2)m(v_(0))^(2)`